Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
         int n = sc.nextInt();
         double arr[] = new double[n];
         for(int i=0;i<n;i++) arr[i] = sc.nextDouble();
         int m = (n+1)/2;
        double dp[][] = new double[n+1][m+1];
        for(int i=1;i<=n;i++) dp[i][0] = 1;
        for(int j=1;j<=m;j++) dp[0][j] = 0;
        dp[0][0] = 1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                dp[i][j] =  arr[i-1] * dp[i-1][j-1] + (1 - arr[i-1]) * dp[i-1][j];
            }
        }

        System.out.print(dp[n][m]);
    }
}